Simple Queueing Models

Transcription

1 CHAPTER 3 Simple Queueing Models We will start with the so-called M/M/ system, where arrivals occur to a single server equipped with an infinite buffer for jobs to wait in. Arriving jobs are served in a FCFS manner, with processing times being exponentially distributed. We start by examining the arrival process in more detail. Using a different viewpoint, we will arrive at the exponential assumption on interarrival times.. The Poisson Process Here we examine the process A(t), which gives the number of arrivals to a system at time t. The Poisson process is a purely random stochastic process, in that we start from the assumptions that () arrivals are independent of one another (2) any one interval of time is no more or less likely to have a particular number of arrivals than one another We develop a model for the Poisson process by dividing the time axis into small amounts, of length t. The probability of an arrival is proportional to the length of the segment, with proportionality constant λ, where λ has the natural interpretation of being the arrival rate. We assume that t is small enough so that P (exactly arrival in (t, t + t]) λ t P (no arrivals in (t, t + t]) λ t P (more than arrival in (t, t + t]) 0 where these approximations become exact as t goes to zero. If we let t 0, the resulting process is a Poisson process. Before we move on to the analysis, this model has the following useful features. If we take a Poisson process and split it by taking each arrival and with some probability p i having it join a new arrival stream i, where i =,..., N and N i= p i =, then we see that for the ith arrival stream, P (exactly arrival in (t, t + t]) p i λ t and thus the i th arrival stream is itself a Poisson process. Similarly, if we combine together a number of independent Poisson arrival streams, we get a Poisson process, as if we take N such arrival streams, 23

2 24 3. SIMPLE QUEUEING MODELS with rates λ,..., λ N, then for the combined stream, P (exactly arrival in (t, t + t]) (λ + λ N ) t. The original application of the Poisson process was to model the arrival of telephone calls to a telephone exchange. The reasoning that such a model was valid was that it was felt that a Poisson process was a good model for the use of each phone, and then the aggregate is also a Poisson process (for reasons given above). Note that if for example, it were true that once an individual call was made, it becomes more likely that subsequent calls are initiated in a short period of time, then a Poisson process would not be appropriate. We will discuss more on modelling issues later, but the Poisson process is quite often applicable and it has the benefit of being analytically tractable. To analyze the behaviour, we need the following definitions: p n (t) = P (A(t) = n) p i,j ( t) = P (A(t + t) = j A(t) = i) With these definitions, we can write Rearranging: p n (t + t) = p n (t)p n,n ( t) + p n (t)p n,n ( t), n p 0 (t + t) = p 0 (t)( λ t) p n (t + t) p n (t) t p 0 (t + t) p 0 (t) t If we let t 0, = λp n (t) + λp n (t), n = λp 0 (t) d dt p n(t) = λp n (t) + λp n (t), n d dt p 0(t) = λp 0 (t) Can we solve these? Certainly for p 0 (t), where using the initial condition p 0 (0) =, we get p 0 (t) = e λt. So, d dt p (t) = λp (t) + λe λt and p (0) = 0 combine to give Continuing, p (t) = λte λt. d dt p 2(t) = λp 2 (t) + λ 2 te λt

3 and p 2 (0) = 0 combine to give 2. THE M/M/ QUEUE 25 p 2 (t) = λ2 t 2 2 e λt. Once we see p 3 (t) = λ3 t 3 6 e λt, we can guess that (5) p n (t) = (λt)n e λt n! and indeed, a quick proof by induction shows that this is true. Example. A telephone exchange receives 00 calls per minute on average, according to a Poisson process. What is the probability that exactly one call is received in an interval of five seconds? The question simply asks for p (/2), where time is measured in minutes. So, ( ) 00 p (/2) = e 00/2 = As (5) is a Poisson distribution with parameter λt, we can use our knowledge of the Poisson distribution to get E[A(t)] = λt V ar(a(t)) = λt The mean in particular should be no surprise, given that λ is the arrival rate. An interarrival time, X, can be described by noting that P (X > t) = p 0 (t) P (X t) = p 0 (t) = e λt which means that X is exponentially distributed, with rate λ. Remember, this has the memoryless property, which also should be no surprise given the independence assumptions that were originally made. 2. The M/M/ queue We need to add to the Poisson process the processing times. As stated earlier, we assume that these are exponentially distributed with rate. In the original work in this area, it was found to be a good model for the length of telephone conversations. It is often used elsewhere, either because it is a decent model or because it can make problems more tractable. We will see later that it is not good for things like file transfers.

4 26 3. SIMPLE QUEUEING MODELS If we try to do a similar derivation as we did for the Poisson process, defining p n (t) = P (n jobs in the system at time t), we see P (exactly arrival in (t, t + t]) λ t P (no arrivals in (t, t + t]) λ t P (exactly departure in (t, t + t], given server busy at time t) t P (no departures in (t, t + t], given server busy at time t) t From these, we can derive d dt p n(t) = (λ + )p n (t) + λp n (t) + p n+ (t), n d dt p 0(t) = λp 0 (t) + p (t). Unfortunately, these equations are difficult to solve, as they are highly coupled. For the Poisson process we had a starting point, here we do not. The solution involves Bessel functions, so while we get a solution, it needs to be evaluated numerically. Instead, we will look for something that can provide a little more insight. We will evaluate lim t p n (t). We already have the tools to do this. The number in the system is an appropriate state to develop a CTMC, whose state transition diagram is given in Figure 3.. The difference between this and the reliability models Figure 3.. M/M/ transition rate diagram is that the state space is infinite. However, we can still use (4) and (3) to derive p 0 ( ( ) ) λ n n=0

5 3. LITTLE S LAW 27 We now use the fact that n=0 ρn = /( ρ), 0 ρ <. If we let ρ = λ/, we then have ρ n p 0 ρ and ( ρ)ρ n. We need ρ < for a solution to exist, but this makes perfect sense, as this simply requires that the arrival rate, λ, is less than the processing rate,. If that were not the case, you would expect the system to not work at all, in that you would see the number of jobs waiting grow with time. Note that the probability that the server is busy is ρ. For this reason, ρ is known as the utilization or load. To get a feel for the numbers, let s look at M/M/ systems with ρ = 0., 0.5 or 0.9. We then have, if N is the random variable representing the steady-state number of jobs in the system: ρ = 0. ρ = 0.5 ρ = 0.9 P (0 N 3) P (4 N 7) P (8 N ) P (N 2) The main observation is that in terms of the number in system, things look highly nonlinear in terms of the utilization of the server. Looking at the steady state distribution itself may be too fine grained to be informative. One performance measure that can be derived from the steady state distribution is the mean number in system, L. L = np n ρ = ρ It is fairly plain to see that L is highly nonlinear in ρ. We can also derive that the variance of the number in system is ρ/( ρ) 2, which is even more nonlinear. n=0 3. Little s Law We just finished deriving the mean number in system, but quite often we are interested in performance measures that involve time, such as the mean response time (the mean time from arrival to departure), or the mean waiting time (time from arrival until processing begins). As the steady state distribution does not involve times, we need to use Little s Law, to relate L and the mean response time, W. Here is a sketch of the derivation of Little s Law. We look at the arrival and departure processes of a generic system (it does not need to be M/M/, Little s Law holds in great generality). For

6 28 3. SIMPLE QUEUEING MODELS example, consider the case of Figure 3.2. Here, A(t) is the number of arrivals Figure 3.2. Derivation of Little s Law in (0, t], D(t) is the number of departures in (0, t] and N(t) is the number of jobs in the system at time t, N(t) = A(t) D(t) (assuming that the system is empty at time 0). From these observations we can derive the average arrival rate at time t to be λ t = A(t), t the average time spent in the system for jobs that arrived in (0, t] to be W t = S(t) A(t), and the average number of jobs in the system over (0, t] to be Combining these gives L t = S(t). t L t = S(t) t = W t A(t) t = λ t Wt. Now, we would expect, as t, L t L, λ t λ and W t W, which means L = λw which is known as Little s Law. As an application, we can use Little s Law to calculate the mean waiting time for the M/M/ queue: W = ρ λ ρ = / ρ. We end our study of the M/M/ queue by calculating a few other performance measures of interest. These are L q, the mean queue length and

7 the mean time in queue, W q. 4. M/M//N - FINITE BUFFER SYSTEM 29 L q = = (n )p n n= np n n= n= = L ( p 0 ) = L ρ ρ 2 = ρ Little s Law also holds for L q, W q, so W q = L q λ = ρ/ ρ p n 4. M/M//N - finite buffer system Here, an arriving job finding N jobs already in the system is lost or blocked. Lost jobs do not return at a later time. Here, the birth and death rates are λ n = λ, n = 0,, 2,..., N n =, n =, 2, 3,..., N The state transition diagram is given in Figure 3.3. We have Figure 3.3. State transition diagram for M/M//N p 0, 0 n N ( n λ ) N n=0

8 30 3. SIMPLE QUEUEING MODELS or λ/ (λ/) N+ λ/ (λ/) N+, 0 n N The blocking probability is the probability that the system is full (p N ), so the mean number of jobs lost per unit time is λp N. Some example values are given below, where we fix the buffer size at N = 5. λ/ p N Note, that for the λ/ =.00 entry, L Hopital s rule is required to calculate p N. Next, we fix λ/ = 0.9 and vary the buffer size. N p N M/M/ - pure delay queue In this system, every job is immediately signed a processor on arrival. Another way to think of this is that it is self-service : each jobs is its own server. The state transition diagram is given in Figure 3.4. Solving for the Figure 3.4. State transition diagram for M/M/

9 6. M/M/c - MULTIPLE SERVER SYSTEM 3 steady-state distribution: ( n ) λ p 0 i i= = p 0 n! + ( i= λ i! = e λ/ where the last equality follows from the fact that So, i=0 n! x i i! = ex. ) i e λ/ Note that this is a Poisson distribution with mean L = λ/. Now, by Little s Law, W = / (this can also be derived as the waiting time is simply a processing time). Trivially, L q = W q = M/M/c - multiple server system Here we have a system with multiple identical processors, sharing a single queue. The state transition diagram is given in Figure 3.5. Solving for the Figure 3.5. State transition diagram for M/M/c steady-state distribution yields n! p 0, n c c!c n c p 0, n c

10 32 3. SIMPLE QUEUEING MODELS which yields [ + c n= n! + c! where ρ = λ c. Using L q = n=c (n c)p n we can derive Using Little s Law, [ L q = [ W q = λ/) c λ (c )!(c λ) 2 λ/) c (c )!(c λ) 2 ( ) λ c ( ) ], ρ ] p 0. ] p 0. We then have W = [ + λ/) c ] (c )!(c λ) 2 p 0, and we can again apply Little s Law to get L = λw. One more performance measure of interest for this model is the probability that an arriving job has to wait before going into service (this is also known as the Erlang C formula. The probability of queueing is ( ) c ( ) λ c! ρ n=c + ( ) n ( ) c ( ). c n= λ n! + λ c! ρ 7. M/M/c/c - loss system Here, if a job arrives to find all c servers busy, it is lost. transition diagram is given in Figure 3.6. Here we have The state Figure 3.6. State transition diagram for M/M/c/c ( n ) λ p 0 i i= = p 0, n c, n!

11 8. CLIENT/SERVER MODEL 33 which yields + ( ) n. c n= λ n! The probability that a job is lost (also known as Erlang s B formula) is: p c = n! ( λ ) n + c n= n! ( λ ) n. 8. Client/Server Model Here we examine a system that consists of a number of clients that wish to use a single server, as in Figure 3.7. The system works as follows - each Figure 3.7. Client/Server model client thinks for a period of time that is exponentially distributed with rate λ. After this think time, a request is sent to the server. After the request is processed at the server, the client begins another think time. We assume that there are M clients. Here, we look at a CTMC where the state is the number of jobs at the server (one could repeat the derivation by choosing the state to be the number of clients thinking). We are not keeping track of individual client requests, just the aggregate, so it does not matter what the scheduling policy at the server is (the processing rate is always as long as the server is working). The state transition diagram is given in Figure 3.8. Here, Figure 3.8. Client/Server state transition diagram

12 34 3. SIMPLE QUEUEING MODELS Solving for p 0 gives = ( n ) λ(m i + ) p 0 i= ( ) M! p 0. (M n)! M n=0 ( λ ) n M! (M n)!. 9. Impact of variance - the M/G/ queue So far, we have looked at models where all underlying distributions are exponential. Can we relax this? One important result is the Pollaczek- Khinchin formula, for the M/G/ queueing system. I will not provide a derivation here, if you are interested you can find it in the book by Gross and Harris on reserve in Thode. Suppose that the processing times have mean and variance σs. 2 Then, if we let ρ = λ/, L = ρ + ρ2 + λ 2 σ 2 s 2( ρ) We can then use Little s Law to calculate W. What does this result say? All things being equal, increasing the variance of processing times causes lower system performance. One example of this is to compare the M/D/ (constant processing times) queue with the M/M/ queue. For M/M/, we have L = ρ ρ, while for the M/D/, using the Pollaczek-Khinchin formula (with σs 2 = 0), we have ρ 2 L = ρ + 2( ρ). Comparing L values for the two queues: ρ M/M/ M/D/

13 9. IMPACT OF VARIANCE - THE M/G/ QUEUE 35 We can verify that, as ρ, the difference between the two mean queue lengths are different by a factor of 2. The degradation of performance due to variability is magnified as the load on the queues increases.